dx-1-cos-x-1-sin-x-

Question Number 167586 by cortano1 last updated on 20/Mar/22

λ = \int \frac{dx}{\sqrt{1 + \cos x} + \sqrt{1 + \sin x}} = ?

Commented by MJS_new last updated on 20/Mar/22

1^{st} step t = \tan \frac{x}{2}

2^{nd} step u = t + \sqrt{t^{2} + 1}

I^{'} ve got no time right now \dots

Answered by greogoury55 last updated on 20/Mar/22

\int \frac{\sqrt{1 + \cos x} - \sqrt{1 + \sin x}}{\cos x - \sin x} d x

= \int \frac{\sqrt{2 \cos^{2} (\frac{x}{2})}}{\cos x - \sin x} d x - \int \frac{\sqrt{{(\cos \frac{1}{2} x + \sin \frac{1}{2} x)}^{2}}}{\cos x - \sin x} d x

= \int \frac{\sqrt{2} \cos \frac{x}{2}}{\cos x - \sin x} d x - \int \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos x - \sin x} d x

Answered by MJS_new last updated on 20/Mar/22

\int \frac{d x}{\sqrt{1 + \cos x} + \sqrt{1 + \sin x}} =

[t = \tan \frac{x}{2} \to d x = \frac{2 d t}{t^{2} + 1}]

= 2 \int \frac{d t}{(t + 1 + \sqrt{2}) \sqrt{t^{2} + 1}} =

[u = t + \sqrt{t^{2} + 1} \to d t = \frac{\sqrt{t^{2} + 1}}{u} d u]

= 4 \int \frac{d u}{u^{2} + 2 (1 + \sqrt{2}) u - 1} =

= \sqrt{2 - \sqrt{2}} \ln \frac{u + 1 + \sqrt{2} - \sqrt{4 + 2 \sqrt{2}}}{u + 1 + \sqrt{2} + \sqrt{4 + 2 \sqrt{2}}} = \dots

Commented by peter frank last updated on 21/Mar/22

thank you